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# Count and Say LeetCode Solution

## Problem – Count and Say

The count-and-say sequence is a sequence of digit strings defined by the recursive formula:

• `countAndSay(1) = "1"`
• `countAndSay(n)` is the way you would “say” the digit string from `countAndSay(n-1)`, which is then converted into a different digit string.

To determine how you “say” a digit string, split it into the minimal number of substrings such that each substring contains exactly one unique digit. Then for each substring, say the number of digits, then say the digit. Finally, concatenate every said digit.

For example, the saying and conversion for digit string `"3322251"`:

Given a positive integer `n`, return the `nth` term of the count-and-say sequence.

Example 1:

``````Input: n = 1
Output: "1"
Explanation: This is the base case.``````

Example 2:

``````Input: n = 4
Output: "1211"
Explanation:
countAndSay(1) = "1"
countAndSay(2) = say "1" = one 1 = "11"
countAndSay(3) = say "11" = two 1's = "21"
countAndSay(4) = say "21" = one 2 + one 1 = "12" + "11" = "1211"``````

Constraints:

• `1 <= n <= 30`

### Count and Say LeetCode Solution in Python

``````def countAndSay(self, n):
s = '1'
for _ in range(n - 1):
s = re.sub(r'(.)\1*', lambda m: str(len(m.group(0))) + m.group(1), s)
return s``````

### Count and Say LeetCode Solution in Java

``````public class Solution {
public String countAndSay(int n) {
String s = "1";
for(int i = 1; i < n; i++){
s = countIdx(s);
}
return s;
}

public String countIdx(String s){
StringBuilder sb = new StringBuilder();
char c = s.charAt(0);
int count = 1;
for(int i = 1; i < s.length(); i++){
if(s.charAt(i) == c){
count++;
}
else
{
sb.append(count);
sb.append(c);
c = s.charAt(i);
count = 1;
}
}
sb.append(count);
sb.append(c);
return sb.toString();
}
}
``````

### Count and Say LeetCode Solution in C++

``````string countAndSay(int n) {
if (n == 0) return "";
string res = "1";
while (--n) {
string cur = "";
for (int i = 0; i < res.size(); i++) {
int count = 1;
while ((i + 1 < res.size()) && (res[i] == res[i + 1])){
count++;
i++;
}
cur += to_string(count) + res[i];
}
res = cur;
}
return res;
}
``````
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